Distances & Area, Left/Right/Midpoint Sums

Question

Estimate the area under the graph of

f(x)= 1/x+3

over the interval [1,3] using four approximating rectangles and right endpoints.

Rn= ?

Repeat the approximation using left endpoints.

Ln= ?

Report answers accurate to 4 places.

Answer 1

y=1/x + 3 could mean y=1/(x+3) or (1/x)+3

assuming the latter

integral of (1/x) +3 = lnx +3x

evaluated from -1 to 3

= ln3+9-(-3)

=ln3+6

= about what you should get with rectangular approximation

each rectangle has base = 1 = (1/4)(3---1) = 4/4

right endpoints are 0,1,2 and 3

heights are f(0), f(1), f(2), f(3)

= 1/0 +3, 1/1+3, 1/2+3, 1/3+3

= UD, 4, 3.5, 10/3, UD for UnDefined

this suggests the problem really was meant to be 1/(x+3)

then

f(0), f(1), f(2), f(3)

= 1/3, 1/4, 1/5 and 1/6

rectangle areas sum to (1/3+1/4+1/5+1/6)(1)

= 7/12 + 11/30

= 57/60

integral of 1/(x+3)

= ln(x+3) evaluated -1 to 3

=ln6- ln2= ln2+ln3-ln2

= ln3 =about 1.0986

which is close to 57/60= .9500

left end points are -1,0,1 and 2

heights are f(-1), f(0), f(1) and f(2)

= 1/2, 1/3, 1/4, 1/5

areas sum to

(1/2+1/3+1/4+1/5)

= 5/6 + 9/20

= 77/60= about 1.2833

.9500 < 1.0986 < 1.2833

midpoint sums should be even closer to 1.0986

midpoints are -1/2, 1/2, 3/2, 5/2

heights f(-1/2), f(1/2), f(3/2), f(5/2)

= 2/5, 2/7, 2/9, 2/11

areas sum to (2/5+2/7+2/9+2/11)(1)

= about 1.0898